Hard Heart of Algebra Problem with Solutions

Yesterday I gave you a Level 5 “Heart of Algebra” problem for the SAT to try. Today I will provide a solution for this problem. If you have not yet attempted the problem go back and take a look at it first so you can try it on your own. Here is the link: Hard Heart of Algebra Problem for the SAT

Level 5 – Heart of Algebra

Here is the problem once again followed by several solutions:

3x – 7y = 12
kx + 21y = -35

For which of the following values of k will the system of equations above have no solution?

(A)  9
(B)  3
(C) -3
(D) -9

Solution: The system of equations

ax + by = c
dx + ey = f

has no solution if a/d = b/e ≠ c/f. So we solve the equation 3/k = -7/21.  Cross multiplying yields 63 = -7k  so that k = 63/-7 = -9, choice (D).

Note: In this problem b/e ≠ c/f. Indeed, -7/21 ≠ 12/-35. This guarantees that the system of equations has no solution instead of infinitely many solutions.

* Quick solution: We multiply -7 by -3 to get 21. So we have k = (3)(-3) = -9, choice (D).

Note: The general form of an equation of a line is ax + by = c where a, b, and c are real numbers. More information about this type of equation can be found in New SAT Math Problems arranged by Topic and Difficulty Level.

Feel free to add your own solutions to the comments.

More Hard SAT Math Practice Problems

For many more hard SAT math problems like these, each with several fully explained solutions, check out the Get 800 collection of SAT math books. Click on the picture below for more information about these books.

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